Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
 
public class Main{
  public static int[] dp;
  public static void main(String[]args){
    Scanner sc =  new Scanner(System.in);
    int N = Integer.parseInt(sc.nextLine());
    String[] s = sc.nextLine().split(" ");
    double[] p = new double[N];
    for(int i = 0; i< N; i++){
      p[i] = Double.parseDouble(s[i]);
    }
    
    double[][] dp = new double[N+1][N+1];
    dp[0][0] = 1.0;
    
    for(int i = 0; i < N; i++){
      for(int j = 0; j <= i; j++){
        dp[i+1][j+1] += dp[i][j] * p[i];
        dp[i+1][j] += dp[i][j] * (1.0 - p[i]);
      }
    }
    double ans = 0.0;
    for(int i = (N/2)+1; i <= N; i++){
      ans += dp[N][i];
    }
    System.out.println(ans);
  }
}